A032130 - OEIS

%I #50 Sep 17 2019 03:49:21

%S 2,2,5,15,52,193,765,3143,13323,57670,254040,1134249,5122124,23349966,

%T 107310784,496633774,2312539465,10826481544,50929829953,240616214596,

%U 1141195080020,5431477088428,25933525825389,124185539096075,596268057962349,2869992942831031,13845453533124431,66934180769445444,324218809545624984

%N Shifts left under the "BIK" (reversible, indistinct, unlabeled) transform with a(1) = 2.

%C From _Petros Hadjicostas_, Jan 14 2018: (Start)

%C For this sequence, if (b(n): n>=1) = BIK((a(n): n>=1)), then b(n) = a(n+1) for n>=1.

%C Let A(x) = Sum_{n>=1} a(n)*x^n be the g.f. for this sequence. For an explanation on how to derive the formula BIK(A(x)) = (1/2)*(A(x)/(1-A(x)) + (A(x) + A(x^2))/(1 - A(x^2))) from Bower's formulae in the link below about transforms, see the comments for sequence A001224. (For that sequence, the roles of sequences (a(n): n>=1) and (b(n): n>=1) are reversed.)

%C (End)

%H Andrew Howroyd, <a href="/A032130/b032130.txt">Table of n, a(n) for n = 1..200</a>

%H C. G. Bower, <a href="/transforms2.html">Transforms (2)</a>

%F From _Petros Hadjicostas_, Jan 14 2018: (Start)

%F The sequence (a(n): n>=1) shifts left under the "BIK" (reversible, indistinct, unlabeled) transform with a(1) = 2.

%F G.f.: If A(x) = Sum_{n>=1} a(n)*x^n, then -a(1) + A(x)/x = BIK(A(x)) = (1/2)*(A(x)/(1-A(x)) + (A(x) + A(x^2))/(1-A(x^2))). Here a(1) = 2.

%F In general, if we let a(1) = c, we get:

%F a(2) = c,

%F a(3) = (1/2)*(c + 3)*c,

%F a(4) = (1/2)*(c + 3)*(c + 1)*c,

%F a(5) = (1/2)*(c^3 + 5*c^2 + 10*c + 4)*c,

%F a(6) = (1/4)*(2*c^4 + 15*c^3 + 38*c^2 + 37*c + 8)*c,

%F a(7) = (1/8)*(4*c^4 + 38*c^3 + 103*c^2 + 109*c + 22)*(c + 1)*c,

%F a(8) = (1/8)*(4*c^6 + 56*c^5 + 251*c^4 + 511*c^3 + 499*c^2 + 201*c + 22)*c,

%F and so on. No pattern is apparent in these formulae.

%F (End)

%e From _Petros Hadjicostas_, Jan 14 2018: (Start)

%e According to Bower's theory in the link above, we have boxes of different sizes and colors. The size of a box is determined by the number of balls it can hold. Two boxes of the same size and color are considered identical (indistinct and unlabeled). We place the boxes on a line that can be read in either direction; i.e., we have a reversible line.

%e Here, a(n) = number of colors a box holding n balls can be, while b(n) = number of ways of placing boxes in a line that can be read in either direction when the total number of balls is n.

%e Since a(1) = 2, a(2) = 2, a(3) = 5, a(4) = 15, etc.,  a box with 1 ball can be of 2 colors only, a box with 2 balls can be of 2 colors only, a box with 3 balls can be of 5 colors only, a box with 4 balls can be of 15 colors, and so on.

%e When we have n=3 balls, we have b(3) = a(4) = 15. Here, we consider three cases. In the first case, we have one box holding 3 balls and we have 5 possibilities. In the second case, we have a box with 2 balls and a box with 1 ball, and we have 2 x 2 = 4 possibilities here because the line is reversible (i.e., 21 is considered the same as 12). In the third case, we have three identical boxes each holding 1 ball and we have 6 possibilities (if the colors are a and b, we have the possibilities aaa, aab, aba, bba, bab, and bbb). Thus, b(3) = 5 + 4 + 6 = 15 = a(4).

%e When we have n=4 balls, we have b(4) = a(5) = 52. Here we consider 5 cases: a single box with 4 balls (a(4) = 15 possibilities); a box with 3 balls and a box with 1 ball (a(3) x a(1) = 5 x 2 = 10 possibilities); two identical boxes each with 2 balls (3 possibilities, aa, ab, and bb); a box with 2 balls and two identical boxes each with 1 ball (14 possibilities, see below); and 4 identical boxes each with 1 ball (10 possibilities, aaaa, aaab, aaba, aabb, abba, baab, abab, abbb, babb, bbbb). Hence, b(4) = 15 + 10 + 3 + 14 + 10 = 52 = a(5).

%e We explain the fourth case above in more detail. Here, we have a box with 2 balls and two identical boxes each with 1 ball. Let a and b be the two colors for the 1-ball boxes and A and B be the colors for the 2-ball boxes. Then we have the following 14 cases: Aaa, Aab, Abb, Baa, Bab, Bbb, aAa, aAb, bAb, aBa, aBb, bBb, abA, and abB. Note that Aab = baA <> abA = Aba and abB = Bba <> Bab = baB.

%e (End)

%t m = 30; a[1] = 2; A[_] = 0;

%t Do[A[x_] = x(a[1]+(1/2)(A[x]/(1-A[x])+(A[x]+A[x^2])/(1-A[x^2]))) + O[x]^m // Normal, {m}];

%t CoefficientList[A[x], x] // Rest (* _Jean-François Alcover_, Sep 17 2019 *)

%o (PARI)

%o BIK(p)={(1/(1-p) + (1+p)/subst(1-p, x, x^2))/2}

%o seq(n)={my(p=O(1));for(i=1, n, p=1+BIK(x*p)); Vec(p)} \\ _Andrew Howroyd_, Aug 30 2018

%Y Cf. A001224, A032131.

%Y When a(1) = 1, we get sequence A032128.

%K nonn

%O 1,1

%A _Christian G. Bower_

%E Name edited by _Petros Hadjicostas_, Jan 14 2018

%E a(24)-a(29) from _Petros Hadjicostas_, Jan 14 2018